Problema Solution
I have a bag containing only red balls and green balls. If I remove three red balls from the bag, four-sevenths of the remaining balls are red. If I double the number of red balls and add one green ball, three fourths of the balls in the bag are red. How many of each color ball did I start with?
Answer provided by our tutors
let 'x' represent the total number of balls, then '(4/7)(x-3) + 3' represents the number of red balls and 'x-((4/7)(x-3)+3)' represents the number of green balls
x = red-balls + blue-balls
x = ((4/7)(x-3) + 3) + ((x-(4/7)(x-3)+3))
double the number of red balls and add one green ball
2((4/7)(x-3)+3) + (x-((4/7)(x-3)+3)+1)
three fourths of the above balls are red:
(3/4)(2((4/7)(x-3)+3) + (x-((4/7)(x-3)+3)+1))
which simplifies to:
(3(11x+16))/28
twice the number of red balls are expressed in "(3(11x+16))/28" as compared to the original count of red balls expressed in "((4/7)(x-3) + 3)"
2*((4/7)(x-3) + 3) = (3(11x+16))/28
solving for 'x' we have x=24
number of red balls:
((4/7)(x-3) + 3) = (4/7)(24-3)+3 = 15
number of blue balls:
24-15 = 9
there were 9 blue balls and 15 red balls